Question

If $f\left(\frac{x-4}{x+2}\right)=2x+1,$ where $x\in R-\{1,-2\},$ then$\int f\left(x\right)dx$ is equal to
(where $C$ is constant of integration)

• A. $12{\log }_e|1-x|-3x+C$
• B. $-12{\log }_e|1-x|-3x+C$
• C. $-12{\log }_e|1-x|+3x+C$
• D. $12{\log }_e|1-x|+3x+C$

Answer 1

The correct option is B $-12{\log }_e|1-x|-3x+C$

Suppose, $\frac{x-4}{x+2}=y\Rightarrow x-4=y(x+2)$
$\Rightarrow x(1-y)=2y+4\Rightarrow x=\frac{2y+4}{1-y}$
So, $f\left(y\right)=2\left(\frac{2y+4}{1-y}\right)+1$
Now,
$f\left(x\right)=2\left(\frac{2x+4}{1-x}\right)+1$
$\Rightarrow f\left(x\right)=\frac{3x+9}{1-x}$
$\Rightarrow f\left(x\right)=\frac{3(x+3)}{1-x}$
$\Rightarrow f\left(x\right)=\frac{3(x-1+4)}{1-x}$
$\Rightarrow f\left(x\right)=-3+\frac{12}{1-x}$
$\therefore \int f\left(x\right)dx=-12{\log }_e|1-x|-3x+C$